If each of these substances costs the same amount per...

If each of these substances costs the same amount per kilogram, which substance would be the most cost effective way to lower the freezing point of water? Assume all ionic compounds completely dissociate. KBr $C_6H_{12}O_6$ $CaCl_2$ NaI $CaI_2$

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Transcript

00:01 Hello everyone.

00:02 The freezing point depression formula is change in temperature equals kf times m times i.

00:14 Where kf is the freezing point constant, m is the molality and i is the number of ions per gram.

00:22 For this question, kf and m would not be the basis of comparison because the kf of water is the same regardless of the compound.

00:31 And the molality is going to be the same for the question because it said per kilogram.

00:36 Thus, our basis of comparison would be the eye.

00:40 So we're looking at the ions per gram.

00:42 The compound with the most ions per gram will give the most freezing depression.

00:48 For the first compound, the first compound is equivalent compound and thus would not give ions in solution.

00:55 Thus, the eye for that compound will be zero.

00:57 The next compound is sodium chloride.

01:02 Nacl and it will give two ions in solution the n -a -plus ion and the c -l -minus ion.

01:10 So the ions per gram will be two ions and one mole of nacl is 58 .44 grams.

01:20 So that's 2 divided by 58 .44 grams and that should give us 0 .0342.

01:30 The next compound is kcl, which would also give two ions in solution, the k plus ion and the cl minus iron.

01:42 Now, one mole of kcl is 74 .55 grams...

Question

If each of these substances costs the same amount per kilogram, which substance would be the most cost effective way to lower the freezing point of water? Assume all ionic compounds completely dissociate. KBr $C_6H_{12}O_6$ $CaCl_2$ NaI $CaI_2$

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